C2F4 Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

C2F4 lewis structure has a double bond between the two Carbon atoms (C) and a single bond between the Carbon atom (C) and Fluorine atoms (F). There are 3 lone pairs on all the Fluorine atoms (F).

If you haven’t understood anything from the above image of C2F4 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of C2F4.

So let’s move to the steps of drawing the lewis structure of C2F4.

Steps of drawing C2F4 lewis structure

Step 1: Find the total valence electrons in C2F4 molecule

In order to find the total valence electrons in a C2F4 molecule, first of all you should know the valence electrons present in carbon atom as well as fluorine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of carbon as well as fluorine using a periodic table.

Total valence electrons in C2F4 molecule

→ Valence electrons given by carbon atom:

Carbon is group 14 element on the periodic table. ^[1] Hence the valence electrons present in carbon is 4.

You can see the 4 valence electrons present in the carbon atom as shown in the above image.

→ Valence electrons given by fluorine atom:

Fluorine is group 17 element on the periodic table. ^[2] Hence the valence electron present in fluorine is 7.

You can see the 7 valence electrons present in the fluorine atom as shown in the above image.

Hence, 

Total valence electrons in C2F4 molecule = valence electrons given by 2 carbon atom + valence electrons given by 4 fluorine atoms = 4(2) + 7(4) = 36.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is C2F4 and it contains carbon atoms (C) and fluorine atoms (F).

You can see the electronegativity values of carbon atom (C) and fluorine atom (F) in the above periodic table.

If we compare the electronegativity values of carbon (C) and fluorine (F) then the carbon atom is less electronegative.

So here, the carbon atoms (C) are the center atom and the fluorine atoms (F) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the C2F4 molecule, you have to put the electron pairs between the carbon-carbon atoms and between the carbon-fluorine atoms.

This indicates that these atoms are chemically bonded with each other in a C2F4 molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of C2F4 molecule, you can see that the outer atoms are fluorine atoms.

These outer fluorine atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the C2F4 molecule.

The C2F4 molecule has a total 36 valence electrons and out of these, only 34 valence electrons are used in the above sketch.

So the number of electrons which are left = 36 – 34 = 2.

You have to put these 2 electrons on both the central carbon atoms in the above sketch of C2F4 molecule.

Now let’s proceed to the next step.

Step 5: Check the octet on the central atom. If it does not have octet, then convert the lone pair into a double bond or triple bond.

In this step, you have to check whether the central carbon atoms (C) are stable or not.

In order to check the stability of the central carbon (C) atoms, we have to check whether they are forming an octet or not.

Unfortunately, one of the carbon atoms is not forming an octet here.

Now to make this carbon atom stable, you have to convert the lone pair into a double bond so that the carbon atom can have 8 electrons (i.e octet).

After converting this electron pair into a double bond, the central carbon atom will get 2 more electrons and thus its total electrons will become 8.

You can see from the above picture that both the carbon atoms are forming an octet.

And hence these carbon atoms are stable.

Now let’s proceed to the final step to check whether the lewis structure of C2F4 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of C2F4.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on carbon (C) atoms as well as fluorine (F) atoms present in the C2F4 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of C2F4 molecule in the image given below.

For Carbon (C) atom:
Valence electrons = 4 (because carbon is in group 14)
Bonding electrons = 8
Nonbonding electrons = 0

For Fluorine (F) atom:
Valence electrons = 7 (because fluorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

From the above calculations of formal charge, you can see that the carbon (C) atoms as well as fluorine (F) atoms have a “zero” formal charge.

This indicates that the above lewis structure of C2F4 is stable and there is no further change in the above structure of C2F4.

In the above lewis dot structure of C2F4, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of C2F4.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

About author

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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